RestTemplate 401 获取错误信息的处理方案

Shun

这篇文章主要介绍了RestTemplate 401 获取错误信息的处理方案，具有很好的参考价值，希望对大家有所帮助。如有错误或未考虑完全的地方，望不吝赐教
目录

• RestTemplate 401错误
  
  
  
  • 异常处理
    
  • 判断是否异常
    
  
  
• RestTemplate通过对象传参，response的body为空讨论
  
  
  
  • 代码复现
    
  • 解决办法一：实体类转成普通类
    
  • 解决办法二：添加注解
    
  
  

RestTemplate 401错误
调用第三方api 若是服务返回状态码不为200，默认会执行DefaultResponseErrorHandler

异常处理

@Override
 public void handleError(ClientHttpResponse response) throws IOException {
  HttpStatus statusCode = getHttpStatusCode(response);
  switch (statusCode.series()) {
   case CLIENT_ERROR:
  throw new HttpClientErrorException(statusCode, response.getStatusText(),
    response.getHeaders(), getResponseBody(response), getCharset(response));
   case SERVER_ERROR:
  throw new HttpServerErrorException(statusCode, response.getStatusText(),
    response.getHeaders(), getResponseBody(response), getCharset(response));
   default:
  throw new RestClientException("Unknown status code [" + statusCode + "]");
  }
 }

判断是否异常

protected boolean hasError(HttpStatus statusCode) {
 return (statusCode.series() == HttpStatus.Series.CLIENT_ERROR ||
   statusCode.series() == HttpStatus.Series.SERVER_ERROR);
}

通常会直接已异常形势抛出，若不特殊处理无法获取返回提示信息。
需要捕捉HttpClientErrorException 异常，则可获取返回信息

try{
    ......
  }catch (HttpClientErrorException e) {
        String resBody = e.getResponseBodyAsString();
        log.info("客户端异常返回:{}", resBody);
        return new ResponseEntity<>(JSON.parseObject(resBody, res), e.getStatusCode());
      }

一开始我这样写，死活返回的都是null
原来跟我设置的requestFactory有关
采用SimpleClientHttpRequestFactory 无法获取提示
需要换成 HttpComponentsClientHttpRequestFactory

RestTemplate通过对象传参，response的body为空讨论

代码复现
实体类

@Entity
@Table(name = "a",schema = "a")
@JsonIgnoreProperties(value = {"a"})
@Setter
@Generated
public class C {
  @Id
  @GeneratedValue
  private Integer id;
  @Column(name = "diseaseName",length = 255,nullable = false,unique = true)
  private String diseaseName;
  @Column(name = "description",length = 255,nullable = false,unique = true)
  private String description;
  @Column(name = "department",length = 255,nullable = false,unique = true)
  private String department;
}
controller
@ResponseBody
  @RequestMapping(value = "",method = RequestMethod.POST)
  public Response APIcreate(@RequestBody C c) {
    String json = JSONUtil.toJSONString(c);
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON_UTF8);
    HttpEntity<String> entity = new HttpEntity<>(json, headers);
    String url = "http://localhost:3001/c";
    ResponseEntity<Commondisease> responseEntity = restTemplate.postForEntity(url, entity, C.class);
    return new ResponseData(ExceptionMsg.SUCCESS, responseEntity);
  }

返回结果截图：

返回结果为空的讨论：返回的C类是jpa封装后的类，即使通过json工具，也无法转换成功

解决办法一：实体类转成普通类

import lombok.AllArgsConstructor;
import lombok.Data;
import lombok.NoArgsConstructor; 
@Data
@AllArgsConstructor
@NoArgsConstructor
public class C { 
  private Integer id; 
  private String diseaseName; 
  private String description; 
  private String department; 
}
 @ResponseBody
  @RequestMapping(value = "",method = RequestMethod.POST)
  public Response APIcreate(@RequestBody C c) {
    //C c = new Commondisease(1,"zhangsan","11","2222");
    String json = JSONUtil.toJSONString(c);
    HttpHeaders headers = new HttpHeaders();
    headers.setContentType(MediaType.APPLICATION_JSON_UTF8);
    HttpEntity<String> entity = new HttpEntity<>(json, headers);
    String url = "http://localhost:3001/c/";
    ResponseEntity<Commondisease> responseEntity =     restTemplate.postForEntity(url,entity,C.class);
    return new ResponseData(ExceptionMsg.SUCCESS,responseEntity);
}

返回成功

解决办法二：添加注解
@Data
以上为个人经验，希望能给大家一个参考，也希望大家多多支持CodeAE代码之家。
原文链接：https://blog.csdn.net/xuyw10000/article/details/88790391